Solutions to all 6 exercises of Chapter 14, Semiconductor Electronics: Materials, Devices and Simple Circuits (NCERT Class 12 Physics Part II, 2026-27): majority and minority carriers in n-type and p-type semiconductors, band gaps of carbon, silicon and germanium, diffusion in a p-n junction, forward bias, and output frequency of half-wave and full-wave rectifiers. All 6 questions are answered, with the key answer highlighted.
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n-type: pentavalent dopant (P, As), electrons are majority carriers. p-type: trivalent dopant (B, Al), holes are majority carriers. Forward bias lowers the potential barrier; reverse bias raises it. Band gaps: C (diamond) about 5.4 eV, Si about 1.1 eV, Ge about 0.7 eV.
In an n-type silicon, which of the following statement is true: (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants.
Solution
n-type silicon is made by adding a pentavalent impurity (such as phosphorus). Each dopant atom gives an extra free electron, so electrons are the majority carriers and holes the minority carriers. Only statement (c) matches.
Which of the statements given in Exercise 14.1 is true for p-type semiconductors?
Solution
p-type material is made by adding a trivalent impurity (such as boron). Each dopant atom creates a hole, so holes are the majority carriers. Statement (d) is true.
Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which of the following statements is true? (a) (Eg)Si<(Eg)Ge<(Eg)C (b) (Eg)C<(Eg)Ge>(Eg)Si (c) (Eg)C>(Eg)Si>(Eg)Ge (d) (Eg)C=(Eg)Si=(Eg)Ge
Solution
Carbon (diamond) has the largest band gap (about 5.4 eV, an insulator), silicon about 1.1 eV and germanium about 0.7 eV. Going down the group, the band gap decreases.
In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a) free electrons in the n-region attract them. (b) they move across the junction by the potential difference. (c) hole concentration in p-region is more as compared to n-region. (d) All the above.
Solution
Diffusion always happens from a region of higher concentration to lower concentration. There are many more holes in the p-region than in the n-region, so holes diffuse into the n-region. (The potential difference across the junction actually opposes this flow.)
When a forward bias is applied to a p-n junction, it (a) raises the potential barrier. (b) reduces the majority carrier current to zero. (c) lowers the potential barrier. (d) None of the above.
Solution
In forward bias the applied voltage opposes the junction's built-in potential, so the barrier is lowered, the depletion layer narrows, and a large majority-carrier current flows.
In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency?
Solution
Half-wave rectifier: only one half of each input cycle passes, so the output has one pulse per cycle: 50 Hz.
Full-wave rectifier: both halves of each cycle give a pulse in the same direction, so there are two pulses per input cycle: 100 Hz.
Half-wave: 50 Hz. Full-wave: 100 Hz.
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